What is the extraneous solution to these equations? $\dfrac{x^2 - 11}{x + 2} = \dfrac{-11x - 29}{x + 2}$
Answer: Multiply both sides by $x + 2$ $ \dfrac{x^2 - 11}{x + 2} (x + 2) = \dfrac{-11x - 29}{x + 2} (x + 2)$ $ x^2 - 11 = -11x - 29$ Subtract $-11x - 29$ from both sides: $ x^2 - 11 - (-11x - 29) = -11x - 29 - (-11x - 29)$ $ x^2 - 11 + 11x + 29 = 0$ $ x^2 + 18 + 11x = 0$ Factor the expression: $ (x + 9)(x + 2) = 0$ Therefore $x = -9$ or $x = -2$ At $x = -2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -2$, it is an extraneous solution.